Quantitative Aptitude

Algebra

3,715 Questions
All Polynomial Roots and Coefficients 2451 Zero-Product Root Finding 667 Linear System Manipulation 162 Polynomial Theorems and Reduction 85 Algebraic Identity Simplification 82 Surds and Indices 70 Modulus and Absolute Value 54 Algebraic Expression Substitution 46 Algebraic Fraction Simplification 33 Logarithmic Equations 25 Repeated Series Sum 12 Linear Equation Word Problems 9

Algebra Questions

Multiple choice

Set values of $K$ for which roots of the quadratic $x^2 - (2K - 1) x + K(K - 1) = 0$ are
  1. both less than $2$ is $K \in (2, \infty)$
  2. of opposite sign is $K \in (-\infty, 0) \cup (1, \infty)$
  3. of same sign is $K \in (-\infty, 0) \cup (1, \infty)$
  4. both greater than $2$ is $K \in (2, \infty)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Roots of x^2 - (2K-1)x + K(K-1) = 0 are (2K-1 +/- sqrt((2K-1)^2 - 4K(K-1))) / 2. Discriminant = 4K^2 - 4K + 1 - 4K^2 + 4K = 1. Roots are (2K-1+1)/2 = K and (2K-1-1)/2 = K-1. For same sign, their product K(K-1) > 0. This holds when K > 1 or K < 0.

Multiple choice

Let $a, b, c, d$ be distinct real numbers and $a$ and $b$ are the roots of quadratic equation ${ x }^{ 2 } - 2cx - 5d = 0$. If $c$ and $d$ are the roots of the quadratic equation ${ x }^{ 2 } - 2ax - 5b = 0$ .then find the numerical values of $a + b + c + d$.
  1. $30$
  2. $15$
  3. $10$
  4. $60$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

a+b = 2c, ab = -5d. c+d = 2a, cd = -5b. Subtracting: (a+b)-(c+d) = 2(c-a) => 3a-3c = d-b. Also ab-cd = -5(d-b). Substituting: ab-cd = -5(3a-3c) = 15c-15a. Solving these symmetric equations leads to a+b+c+d = 30.

Multiple choice

lf the roots of the equation $\mathrm{a}\mathrm{x}^{2}+\mathrm{b}\mathrm{x}+\mathrm{c}=0$ are the reciprocals of the roots of the equation $\mathrm{p}\mathrm{x}^{2}+\mathrm{q}\mathrm{x}+\mathrm{r}=0$, then
  1. $\mathrm{a}\mathrm{c}\mathrm{q}^{2}=\mathrm{b}^{2}\mathrm{p}\mathrm{r}$
  2. $\mathrm{a}\mathrm{c}=\mathrm{p}\mathrm{r}$
  3. $\mathrm{b}^{2}\mathrm{a}\mathrm{c}=\mathrm{q}^{2}\mathrm{p}\mathrm{r}$
  4. $\mathrm{a}\mathrm{b}=\mathrm{p}\mathrm{q}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If roots of ax^2+bx+c=0 are reciprocals of roots of px^2+qx+r=0, then the equation with roots 1/alpha and 1/beta is obtained by swapping coefficients: cx^2+bx+a=0. Comparing this to px^2+qx+r=0, we have c/p = b/q = a/r. From b/q = a/r and b/q = c/p, we derive relations. Specifically, (b/q)^2 = (a/r)*(c/p) = ac/pr, which leads to b^2pr = q^2ac.

Multiple choice

lf $\alpha$ and $\beta$ are the roots of $\mathrm{x}^{2}+\mathrm{p}\mathrm{x}+\mathrm{q}=0$ and $\alpha^{4},\ \beta^{4}$ are the roots of $\mathrm{x}^{2}-\mathrm{r}\mathrm{x}+\mathrm{s}=0$, then the equation $\mathrm{x}^{2}-4\mathrm{q}\mathrm{x}+(2\mathrm{q}^{2}-\mathrm{r})=0$ has

  1. two real roots

  2. two negative roots

  3. two positive roots

  4. one positive and one negative root
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given roots of x^2+px+q=0 are a, b. Then a+b = -p, ab = q. Roots of x^2-rx+s=0 are a^4, b^4. Then a^4+b^4 = r, a^4b^4 = s = q^4. The equation x^2 - 4qx + (2q^2 - r) = 0 has discriminant D = 16q^2 - 4(2q^2 - r) = 8q^2 + 4r. Since r = a^4+b^4 > 0, D > 0. Product of roots is 2q^2 - r = 2q^2 - (a^4+b^4). Since a^4+b^4 > 2(a^2b^2) = 2q^2, the product is negative, implying roots have opposite signs.

Multiple choice

Ramesh and Mahesh solve an equation. In solving Ramesh commits a mistake in constant term and find the roots are $8$ and $2$ Mahesh commits a mistake in the coefficient of x and find roots $-9$ and $-1$. The correct roots are
  1. $-8,2$
  2. $9,1$
  3. $9,-1$
  4. $-8,-2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Ramesh got roots 8 and 2, so the equation is (x-8)(x-2) = x^2 - 10x + 16. He made a mistake in the constant term, so the coefficient of x (-10) is correct. Mahesh got roots -9 and -1, so the equation is (x+9)(x+1) = x^2 + 10x + 9. He made a mistake in the coefficient of x, so the constant term (9) is correct. The correct equation is x^2 - 10x + 9 = 0. Factoring gives (x-9)(x-1) = 0, so roots are 9 and 1.

Multiple choice

If $\alpha, \beta$ are the roots of the equation $ax^{2} + bx + c = 0$, then $log (a - bx + cx^{2})$ is equal to
  1. $\log a + (\alpha + \beta) x + \dfrac {\alpha^{2} + \beta^{2}}{2} x^{2} + \dfrac {\alpha^{3} + \beta^{3}}{3}$
  2. $\log a + (\alpha + \beta) x - \left (\dfrac {\alpha^{2} + \beta^{2}}{2}\right ) \cdot x^{2} + \left (\dfrac {\alpha^{3} + \beta^{3}}{3}\right )x^{3}$
  3. $\log a - (\alpha + \beta) x - \left (\dfrac {\alpha^{2} + \beta^{2}}{2}\right ) x^{2} - \left (\dfrac {\alpha^{3} + \beta^{3}}{3}\right )x^{3}$

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the Taylor expansion for log(1-u) = -u - u^2/2 - u^3/3..., the expression log(a(1 - (b/a)x + (c/a)x^2)) = log(a) + log(1 - (b/a)x + (c/a)x^2). Given roots alpha, beta, (b/a) = alpha+beta and (c/a) = alpha*beta. The expansion matches option B.

Multiple choice

If $c\neq 0$ and $\alpha, \beta$ are the roots of the quadratic equation $ax^2 + bx +c=0$, then value of $\displaystyle \frac{a\alpha^2 + c}{a\alpha +b}+\frac{a\beta^2+ c}{a\beta+b}$ is
  1. $\displaystyle \frac{b(b^2-2ac)}{4a}$
  2. $\displaystyle \frac{(b^2-2ac)}{2a}$
  3. $\displaystyle \frac{b(b^2-2ac)}{a^2c}$
  4. $\displaystyle \frac{b(b^2-2ac)}{ac}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given ax^2 + bx + c = 0, roots alpha, beta satisfy a*alpha^2 + b*alpha + c = 0, so a*alpha^2 + c = -b*alpha. The expression becomes (-b*alpha)/(a*alpha + b) + (-b*beta)/(a*beta + b). This simplifies using symmetric sums of roots.

Multiple choice

If $ax^{2} + bx + c = 0$ has imaginary roots and $a - b + c > 0$, then the set of points $(x, y)$ satisfying the equation $\left |a\left (x^{2} + \dfrac {y}{a}\right ) + (b + 1) x + c\right | = |ax^{2} + bx + c| + |x + y|$ consists of the region in the xy-plane which is

  1. On or above the bisector of I and III quadrant

  2. On or above the bisector of II and IV quadrant

  3. On or below the bisector of I and III quadrant

  4. On or below the bisector of II and IV quadrant
Reveal answer Fill a bubble to check yourself
A Correct answer